You drop a ball from a window on an upper floor of a building and it is caught b

Question

You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 18.1 m. A) at what time do they pass each other? B) at what location do they pass each other relative to the window? You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 18.1 m. A) at what time do they pass each other? B) at what location do they pass each other relative to the window? You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 18.1 m. A) at what time do they pass each other? B) at what location do they pass each other relative to the window? You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 18.1 m. A) at what time do they pass each other? B) at what location do they pass each other relative to the window?

Explanation / Answer

(A) as ball is dropped,

v0 = 0

a = - 9.8 m/s^2

y = 0 - 18.1 = - 18.1 m

Applying vf^2 - vi^2 =2 a d

vf^2 - 0 = 2(-9.8)(18.1)

vf = 18.8 m/s

suppose ball meet after time t,

then d1 + d2 = 18.1 m

(9.8 t^2 /2 ) + (18.8 t - 9.8 t^2 /2 ) = 18.1

t = 18.1 / 18.8 = 0.96 sec ......Ans

(B) from window,

d = 9.8 x 0.96^2 / 2 = 4.5 m below the window

  

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